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Ludmilka [50]
3 years ago
15

Does gas have its own density

Chemistry
2 answers:
Leokris [45]3 years ago
6 0

Answer:

The identity does not matter because the variables of Boyle's law do not identify the gas.

Explanation:

The ideal gas law confirms that 22.4 L equals 1 mol.

ikadub [295]3 years ago
5 0

Explanation:

yes, it does. A gas with a small molar mass will have a lower density than a gas with a large molar mass.

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(01.04 MC)
iogann1982 [59]
I turned in this exact assignment today haha

the blood vessels dilate to draw body heat away from the body and towards the surface of the skin
sweat glands release sweat, when sweat evaporates it releases heat
6 0
3 years ago
One question please help!
Agata [3.3K]
<span>1 trial : you have nothing to compare the result with - you don't know if it's a mistake.
2 trials : you can compare results - if very different, one may have gone wrong, but which one?
3 trials : if 2 results are close and 3rd far away, 3rd probably unreliable and can be rejected.

******************************

First calculate the enthalpy of fusion. M, C and m,c = mass and specific heat of calorimeter and water; n, L = mass and heat of fusion of ice; T = temperature fall.

L = (mc+MC)T/n.

c=4.18 J/gK. I assume calorimeter was copper, so C=0.385 J/gK.

1. M = 409g, m = 45g. T = 22c, n = 14g
L = (45*4.18+409*0.385)*22/14 = 543.0 J/g.

2. M = 409g, m = 49g, T = 20c, n = 13g
L = (49*4.18+409*0.385)*20/13 = 557.4 J/g.

3. M = 409g, m = 54g, T = 20c, n = 14g
L = (54*4.18+409*0.385)*20/14 = 547.4 J/g.

(i) Estimate error in L from spread of 3 results.
Average L = 549.3 J/g.
average of squared differences (variance) = (6.236^2+8.095^2+1.859^2)/3 = 35.96
standard deviation = 5.9964
standard error = SD/(N-1) = 5.9964/2 = 3 J/g approx.

% error = 3/547 x 100% = 0.5%.

(ii) Estimate error in L from accuracy of measurements:
error in masses = +/-0.5g
error in T = +/-0.5c

For Trial 3
M = 409g, error = 0.5g
m = 463-409, error = sqrt(0.5^2+0.5^2) = 0.5*sqrt(2)
n =(516-463)-(448-409)=14, error = 0.5*sqrt(4) = 1.0g
K = (mc+MC)=383, error = sqrt[2*(0.5*4.18)^2+(0.5*0.385)^2] = 2.962

L = K*T/n
% errors are
K: 3/383 x 100% = 0.77
T: 0.5/20 x 100% = 2.5
n: 1.0/14 x 100% = 7.14

% errors in K and T are << error in n, so we can ignore them.
% error in L = same as in n = 7% x 547.4 = 40 (always round final error to 1 sig fig).

*************************************

The result is (i) L= 549 +/- 3 J/g or (ii) L = 550 +/- 40 J/g.
Both are very far above accepted figure of 334 J/g, so there is at least one systematic error in the experiment or the calculations.
eg calorimeter may not be copper, so C is not 0.385 J/gK. (If it was polystyrene, which absorbs/ transmits little heat, the effective value of C would be very low, reducing L.)
Using +/- 40 is probably best (more cautious).
However, the spread in the actual results is much smaller; try to explain this discrepancy - eg
* measurements were "fiddled" to get better results; other Trials were made but only best 3 were chosen.
* measurements were more accurate than I assumed (eg masses to nearest 0.1g but rounded to 1g when written down).

Other sources of error:
L=(mc+MC)T/n is too high, so n (ice melted) may be too small, or T (temp fall) too high - why?
* it is suspicious that all final temperatures were 0c - was this actually measured or just guessed? a higher final temp would reduce L.
* we have assumed initial and final temperature of ice was 0c, it may actually have been colder, so less ice would melt - this could explain small values of n
* some water might have been left in container when unmelted ice was weighed (eg clinging to ice) - again this could explain small n;
* poor insulation - heat gained from surroundings, melting more ice, increasing n - but this would reduce measured L below 334 J/g not increase it.
* calorimeter still cold from last trial when next one started, not given time to reach same temperature as water - this would reduce n.
Hope This Helps :)
</span>
3 0
3 years ago
An adiabatic throttling valve expands an ideal gas. The internal energy of the ideal gas?
just olya [345]

The internal energy of the ideal gas is zero

The change in internal energy  for an isothermal process is zero.

An ideal gas has no interactions between particles, therefore no intermolecular forces.

pressure change at constant temperature does not change the internal energy.

Adiabatic throttling expansion has less work done and lower heat flow.

That lower the internal energy.

The temperature decreases during the adiabatic expansion

Hence the internal energy of the ideal gas is zero

Learn more about the ideal gas on

brainly.com/question/17136449

#SPJ4

3 0
2 years ago
Describe why the overgrowth of algae in eutrophication is a bad thing.
Tanya [424]

Answer:

The overgrowth of algae consumes oxygen and blocks sunlight from underwater plants. When the algae eventually dies, the oxygen in the water is consumed. The lack of oxygen makes it impossible for aquatic life to survive

Explanation:

hope this helps have a good rest of your afternoon :) ❤

7 0
2 years ago
Read 2 more answers
The elementary gas-phase reaction takes place isobarically and isothermally in a PFR where 63.2% conversion is achieved. The fee
a_sh-v [17]

Answer:

Explanation:

check below for explanation in the attached files.

6 0
3 years ago
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